a) Skorzystamy ze wzoru na różnicę kwadratów:

$w\left(x,y\right)={\left(2x-y\right)}^2-{\left(2x+y\right)}^2=\left(2x-y-\left(2x+y\right)\right)\left(2x-y+2x+y\right)=\left(2x-y-2x-y\right)\cdot 4x=-2y\cdot 4x=-8xy$  

$\text{Dla}x=\sqrt{6},y=\sqrt{3},-8\cdot \sqrt{6}\cdot \sqrt{3}=-8\cdot 3\cdot \sqrt{2}=-24\sqrt{2}$  

b) Skorzystamy ze wzoru na różnicę kwadratów:

$w\left(x,y\right)={\left(x+5y\right)}^2-{\left(5y-x\right)}^2=\left(x+5y-\left(5y-x\right)\right)\left(x+5y+5y-x\right)=\left(x+5y-5y+x\right)\cdot 10y=2x\cdot 10y=20xy$  

$\text{Dla}x=\sqrt[3]{3},y=\sqrt[3]{9},20\cdot \sqrt[3]{3}\cdot \sqrt[3]{9}=20\cdot \sqrt[3]{27}=20\cdot 3=60$   

c) Skorzystamy ze wzoru na kwadrat różnicy:

$w\left(x,y\right)={\left(12x-15y\right)}^2-20\left(7x^2+10y^2-17xy\right)=144x^2-360xy+225y^2-140x^2-200y^2+340xy=4x^2-20xy+25y^2={\left(2x-5y\right)}^2$  

$\text{Dla}x=5,y=2,{\left(2\cdot 5-5\cdot 2\right)}^2=0$